solved examples based on progression

7/29/2019 Solved Examples Based on Progression

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Solved Examples based on Progression Part 1

Solved Examples

Example 1:

Find the sum of the series1 3 + 5 7 + 9 11 + upto n terms.

Solution :

This problem can be solved by different approachesm of course some one is

tricky, some one is little bit lengthy. Now you have to judge which one issuitable for your.

Method I:

In the problem, it is not mentioned that n is even or n is odd. So, take

the different case.

Case I:

When n is even

i.e. n = 2m

Now, given series can be broken into two series like(1 + 5 + 9 + 13 +m terms)= m/2 [2 + (m 1)4] m/2 [2 3 + (m 1)4]= m/2 (4) = 2m= n

Case II:

When n is odd

i.e. n = 2m + 1.[1 + 5 + 9 +(m + 1) terms] [3 + 7 + 11 + m terms]= m+1/2 [2 + 4m] m/2 [6 + (m1)4]= m/2 [2 + 4m 6 4m + 4] + 1 + 2m= 2m + 1= n

Method 2:


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1 3 + 5 7 + 9 upto n terms

Case I:

When n is even i.e. n = 2mTake two terms of given series at a time then above series becomes

(2) + (2) + (2) + upto m times.So, sum of series = 2m= n

Case II:

When n is odd i.e. n = 2m + 11 + (2 + 2 + 2 + upto m times)= 1 + 2 m= n.

Method 3:

1 3 + 5 7 + 9 11 + upton termsThis series is also written as

1 + 3.(1) + 5.(1)2 + 7.(1)3 + 9.(1)4+ upto n terms1 + [3.(1) + 5.(1)2 + 7.(1)3 + 9.(1)4+ upto (n1) terms]

And seeing clearly, it is arithmetic geometric series is and commondifference is 2 and the first term of geometric series is (3) and common

ratio is (1).Sn = 1 + [3.(1) + 5.(1)2 + 7.(1)3++ (2n1).(2)n1]

Let Sn1 = 3.(1) + 5.(1)2++ (2n1).(1)

n1

Sn1 = 3.(1)2++ (2n3) (1)

n1+ (2n1)(1)

n

Subtracting, we get,

2 Sn1 = 32 [(1) + (1)2++ (1)

n2](2n1)(2)

n

Case I:


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When n is even

2 Sn1 = 3 2n+1 = 2 (n+1)Sn1 = (n+1)Sn = 1 + Sn1 = n

Example 2:

The sum of three consecutive terms in A.P. is 27 and the sum of theirsquares is 293, then find all three terms.

Solution:

If we take three terms of an A.P. like a, a + d. and a + 2dthen a + (a + d) + (a + 2d) = 3(a + d) = 27 anda2 + (a+d)2 + (a+2d)2 = 293In this way calculation is very long, so we take the terms like

, , , + are the required numbers.so (a ) + + ( + b) = 27or 3 = 27 = 9Now ()2+ 2+ (+)2 = 293.or 2 + 22 + 2+ 2 + 2+ a = 293or 32 + 2= 293or 22 = 293 32 = 293 3(92) = 293 243 = 50.or 2 = 25

= +5

So the numbers are 4, 9, 14. Negative values will just reverse the order. (Ans.)

Example 3:

Find the common difference of an A.P. whose first term is 100 and the sumof whose first six terms is five times the sum of the next six terms.

Solution:

Let the common difference is d. Sn = n/2 [2a + (n1)d]Sn = 6/2 [2 100(n1)d] (i)

= 3[200 + 5d]S7.12 = 7thterm+12thterm/2 6(using equation 4)

= 100+6d+100+11d/2 6


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= (200 + 17d) 3 (ii)

According to problem 5 eq. (i) = eq. (ii)So 5 (200 + 7d)3 = 3(200 + 5d)

Or 1000 + 85d = 200 + 5dOr 80 d = 800

So d = 10


Example 4:

The sum of three consecutive numbers in G.P. is 39 and their product is 729

then find all three numbers.

Solution :

Let /, , are required numbersProduct = 3 = 729 = 93 = 9/ + + = 39 (given)(1/ + 1 + ) = 39

Multiplying each term by

9(1 + + 2) = 39Or 92 + 9 + 9 39 = 0Or 92 30 + 9 = 0 = 30+900324/18 = 30+576/18 = 30+24/18 = 54/18, 6/18 = 3, 1/3

So numbers are 3, 9, 27 or 27, 9, 3. (Ans.)

Example 5:

If Sp denote the sum of the series 1 + rp + r2p+ and sp denote

the sum of series 1 rp + r2p (assume |r| < 1) then prove that

Solution:

Sp = 1 + rp + r2p+ (i)


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=1/1rpsp =1/1r

p

(here common ratio is rp) Sp + sp = 1/1rp + 1/1rp = 1+rp+1rr/1(rp)2

= 2/1r2p

= 2 S2p

Example 6:

Prove that 0.423232323 is a rational number

Solution:

This problem could be solved by either using progression or without

progression.Let S = 0.4232232

Method 1:

S = 4/10 + 23/1000 + 23/100000 +...

= 4/10 + 23/103 + 23/105 +...= 4/10 + 23/103 [1 + 1/102 + 1/104 +...]

= 4/10 + 23/103 [1/11/102]

= 4/10 + 23/103

100/99= 4/10 + 23/103 = 419/990

So S is a rational number. (Proved)

Method 2:

1000 S = 423.232323 10 S = 4.232323 ..Subtracting, we get,990 S = 419

S = 419/990

So S is a rational number. (Proved)


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Solved Examples based on Progression

Part 3

Solved Examples based on Progression :

Example:

If a2, b2, c2 are in A.P. show that b+ c, c, c+ a, a + b are in H.P.

Solution:

a2, b2, c2 are in A.P.

So adding ab + bc + ca to each term

a2

+ ab + bc + ca, b2

+ ab + bc + ca. c2

+ ab + bc + ca will also be inA.P.

i.e. a(a + b) + c(b+a), b(a+b) + c(b+a), c(c+b) + a(b+c) are in A.P. i.e. (a+c)(a+b), (b+c)(a+b),(c+a(b+c) are in A.P.

dividing each term by (a+b)(b+c)(c+a)1/b+a, 1/c+a, 1/a+b are in A.P.

So b + c, c+a, a+b are in H.P. (Proved)

Example :

If the AM between a and b is twice as greater as the GM, show thata/b = 7+43

Solution:

A.M. =a/b

G.M. =abWe have givenAM = 2 GM

a+b/2 = 2aba+b4 (ab)1/2 = 0

divide each term by b(a/b) + 1 4 (a/b)1/2 = 0or (a/b)24 a/b + 1 = 0a/b = 4+164/2 = 4+23/2 = 2+Squaring both sides, we get,

a/b = (2+3)2 = 4 + 3 + 43 = 7+43

Example :


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The sequence N of natural numbers is divided into classes as follows1

2 3 45 6 7 8 9

10 11 12 13 14 15 16----------------------------------------------

-------------------------------------------------- Find the sum of the numbers of j-th row

Solution:

Numbers of element in j-th row = (2j1)So number of elements in first j rows

= (2j-1) = 2 j 1= 2 j(j+1)/2 j

= j(j+11)=j2

So number of elements in first (j-1) class = (j-1)2So first element of j-th class = (j1)1 +1So sum of (2j-1) terms of A.P. with first term, (j-1)2Common difference, 11 = (2j1)/2 [2{(j1)2 + 1}+(2j1).1]

= (2j 1) (j2 2j + 2 + j 1)

= (2j 1) (j2

j + 1). (Ans.)

Example :

If a1, a2, , an are in arithmetic progression, where ai > 0 for all I,show that

1/a1+ a2, 1/a2+ a3, ......., 1/an1+ an = n1/a1+ an

Solution:

L.H.S. = 1/a1+ a2, 1/a2+ a3, ......., 1/an1+ an = n1/a1+ an= (a1+a2)/(a1+a2)(a1a2), (a2+a3)/(a2+3)(a2a3)

++ (an1an)/(an1+an)(an1an)= a1a2/a1 a2+ a2a3/a2 + a3+...+ an1+ an/an1 + an

Since a1, a2, a3, , an are in A.P.So a2 a1 = a3 a2= = an an1 = d (say)Adding all terms, we have an a1 = (n1)d (i)So L.H.S. = a1a2/d + a2a3/d +...+ an1an/d


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=[a1a2+ a2a3+...+ an1an]= an+a1/d = ana1/(ana1)/(n1) = (n1)(an

a1)/(an+a1)(ana1) (from I)= n1/an+a1 = R.H.S. (Proved)

Example :

If S be the sum, P the product and R the sum of the reciprocals of n terms ofa G.P.

Solution:

Given S = a + ar + ar2++ arn1

P = a.ar.ar2 arn1

R =1/a + 1/ar + 1/ar2 +...+ 1/arn1

So, S =a(1r)r/1rP = an.rn(n1)/2R =(1/a)(11/rn)/11/rSo L.H.S. = (S/R)r = [a(1rr)/(1r)1/a(11/rn)/(11/r)]

= [a2(1rn)(r1/r)/(1r)(rn1/rr)]n

= [a2

(1rn

)(r1)rn

/r(1r)(1rn

)]n

= [a2

rn1

]n

= [an

rn(n1)/2

]2

= P2 = R.H.S. (Proved)


Example:

Find the sum to n terms of the series

1.3.5 + 3.5.7 + 5.7.9. +

Solution:

Here tn = (2n1)(2n+1)(2n+3)


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Sn = tn= (2n1)(2n+1)(2n+3)= ((2n)212) (2n+3)= (4n21)(2n+3)= (8n3 + 12 n2 2n 3)= 8 n2(n+1)2/4 + 12 n(n+1)(2n+1)/6 2 n(n+1)/2 3n

= n[2n (n+1)2 + 2(n+1)(2n+1)(n+1) 3]= n[2n (n2 + 2n + 1) +2(2n2 + 3n + 1) n 4]= n[2n3 + 4n2 + 2n2 + 4n2 + 6n + 2 n 4]= n[2n3 + 8n2 + 7n 2]

Hence Sn = n[2n3 + 8n2 + 7n 2]

Example:

Find the sum of n terms of the series 1/1.2.3.4 + 1/2.3.4.5 + 1/3.4.5.6 +.........

Solution:

tr = 1/r(r+1)(r+2)(r+3)tr+1 = 1/(r+1)(r+2)(r+3)(r+4)

tr/tr+1 = r+4/rr tr = (r + 4) tr+1

r tr= (r + 1) tr+1 + 3 tr+1

Putting r = 1, 2, , n1adding we get, t1 ntn = 3 [t2 + t2 + t3++ tn]

or 4t1ntn = 3[t1 + t2++ tn] = 3 Sn

Sn = 1/3 [4t1 ntn]=1/3 [1/1.2.3 m/3(n+1)(n+2)(n+3)]= 1/18 1/3(n+1)(n+2)(n+3) (Ans.)

Example:

Sum of the following series to n terms:

1/1.3 + 2/1.3.5 + 3/1.3.5.7 + 4/1.3.5.7.9 +......

Solution:

tm = m/1.3.5.7 ...... (2m+1)= 1/2 [2m+11/1.3.5.7 ...... (2m+1)]=1/2 {1/1.3.5 ...... (2m1) 1/1.3.5 ...... (2m+1)}

m = 1: t1 =1/2 {1 1.3}m = 2: t2 =1/2 {1/1.3 1/1.3.5}m = 3: t1 =1/2 {1/1.3.5 1/1.3.5.7}m = n: tn =1/2 {1/1.3.5..... (2n1) 1/1.3.5......(2n+1)}


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Adding,

Sn = t1 + t2+tn = 1/2 {1 1/1.3.5 ..... (2n+1)}

solved examples based on progression

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